Sum the series : $\frac{1}{9\sqrt11 + 11\sqrt9} +\frac{1}{11\sqrt13 + 13\sqrt11} +\ldots$

Question

$$\frac{1}{9\sqrt11 + 11\sqrt9} + \frac{1}{11\sqrt13 + 13\sqrt11} + \frac{1}{13\sqrt15 + 15\sqrt13} + \ldots + \frac{1}{n\sqrt{n+2} + (n+2)\sqrt{n}} = \frac{1}{9}$$

Find the value of $n$.

I got the summation as $$\sum_{r=1}^n{\frac{1}{(2r+7)\sqrt{2r+9} + (2r+9)\sqrt{2r+7}}} = \sum_{r=1}^n{\frac{1}{\sqrt{2r+7}\sqrt{2r+9}(\sqrt{2r+7} + \sqrt{2r+9})}}$$ but I'm not able to simplify it further. I think I should've expressed it as difference of two summations but I'm not able to figure out which summations. Please help.

Answer 1

$$S_m=\sum_{r=1}^m{\frac{1}{(2r+7)\sqrt{2r+9} + (2r+9)\sqrt{2r+7}}} =\sum_{r=1}^m{\frac{(2r+7)\sqrt{2r+9} - (2r+9)\sqrt{2r+7}}{(2r+7)^2(2r+9)-(2r+9)^2(2r+7)}}$$ $$= \sum_{r=1}^m{\frac{(2r+7)\sqrt{2r+9} - (2r+9)\sqrt{2r+7}}{-2(2r+7)(2r+9)}} = -\frac{1}{2}\sum_{r=1}^m{\frac{\sqrt{2r+9}}{2r+9}} - {\frac{\sqrt{2r+7}}{2r+7}}$$ $$=\frac{1}{2}\sum_{r=1}^m{\frac{1}{\sqrt{2r+7}}} - {\frac{1}{\sqrt{2r+9}}} = \frac{1}{2}\left({\frac{1}{3}} - {\frac{1}{\sqrt{2m+9}}}\right) $$

So $$S_m = \frac{1}{9} \Longleftrightarrow {\frac{1}{\sqrt{2m+9}}} = \frac{1}{9} \Longleftrightarrow \sqrt{2m+9} = 9 \Longleftrightarrow m=36$$

Answer 2

Observe $$\dfrac{1}{9\sqrt{11} + 11\sqrt9}$$

$$ =\dfrac{1}{\sqrt 9 \sqrt {11} }\cdot\dfrac{1}{\sqrt9 + \sqrt{11}}$$

$$ =\dfrac{1}{\sqrt 9 \sqrt {11} }\cdot \dfrac{\sqrt{11} - \sqrt{9}}{2}$$

$$ =\dfrac{1}{2}\cdot \dfrac{\sqrt{11} - \sqrt{9}}{\sqrt 9 \sqrt {11} }$$

$$ =\dfrac{1}{2}\cdot \Big(\dfrac{1}{\sqrt{9}} - \dfrac{1}{\sqrt{11}}\Big)$$

Thus LHS

$$ \dfrac{1}{2}\cdot \Big(\dfrac{1}{\sqrt{9}} - \dfrac{1}{\sqrt{11}}\Big) + \dfrac{1}{2}\cdot \Big(\dfrac{1}{\sqrt{13}} - \dfrac{1}{\sqrt{11}}\Big) + \ldots + \dfrac{1}{2}\cdot \Big(\dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n+2}}\Big)$$

telescopes to

$$ \dfrac{1}{2}\cdot \Big(\dfrac{1}{\sqrt{9}} - \dfrac{1}{\sqrt{n+2}}\Big) = \dfrac{1}{9}$$

$$ \Rightarrow \dfrac{1}{3} - \dfrac{1}{\sqrt{n+2}} = \dfrac{2}{9}$$

$$ \Rightarrow n+2 = 81 $$

$$ \Rightarrow \boxed{n=79} $$

Note The $n$ in your question and the $n$ in your attempt are different. I have found $n$ in your question. Indeed final term is $\frac{1}{\sqrt{81}}$ as in the other answer. The $n$ in your attempt is $36$. Indeed $2\cdot36+9=81$.